Question: Graph this system of equations and solve. $14x+10y = 50$ $-6x-10y = -10$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ Click and drag the points to move the lines.
Solution: Convert the first equation, $14x+10y = 50$ , to slope-intercept form. $y = -\dfrac{7}{5} x + 5$ The y-intercept for the first equation is $5$ , so the first line must pass through the point $(0, 5)$ The slope for the first equation is $-\dfrac{7}{5}$ . Remember that the slope tells you rise over run. So in this case for every $7$ positions you move down (because it's negative) You must also move $5$ positions to the right. $5$ positions to the right. $7$ positions down from $(0, 5)$ is $(5, -2)$ Graph the blue line so it passes through $(0, 5)$ and $(5, -2)$ Convert the second equation, $-6x-10y = -10$ , to slope-intercept form. $y = -\dfrac{3}{5} x + 1$ The y-intercept for the second equation is $1$ , so the second line must pass through the point $(0, 1)$ The slope for the second equation is $-\dfrac{3}{5}$ . Remember that the slope tells you rise over run. So in this case for every $3$ positions you move down (because it's negative) You must also move $5$ positions to the right. $5$ positions to the right. $3$ positions down from $(0, 1)$ is $(5, -2)$ Graph the green line so it passes through $(0, 1)$ and $(5, -2)$ The solution is the point where the two lines intersect. The lines intersect at $(5, -2)$.